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From: Giorgos Keramidas <keramida@ceid.upatras.gr>
To: trashy_bumper@yahoo.com
References: <266527.92897.qm@web110502.mail.gq1.yahoo.com>
Date: Sun, 21 Sep 2008 17:17:27 +0300
In-Reply-To: <266527.92897.qm@web110502.mail.gq1.yahoo.com> (Nash Nipples's
	message of "Sun, 21 Sep 2008 05:57:06 -0700 (PDT)")
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Cc: freebsd-questions@freebsd.org
Subject: Re: Segmentation fault when free
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On Sun, 21 Sep 2008 05:57:06 -0700 (PDT), Nash Nipples <trashy_bumper@yahoo.com> wrote:
> thanks for making it even more clear to me.
> actually what i meant was this:
>
> void function(void){
>   char *p;
>   p = malloc(1);
> }
> int main(void){
>   while (1){
>     function();
>     /* in the end of this function function()
>      * the memory is still allocated
>      * even when the only pointer who knows its address
>      * does not longer exist
>      * which is why we have to free() the memory
>      * during the application runtime
>      * to avoid it from growing to ridiculous size
>      */
>   }
> }

This won't throw SEGV in free() because, well, it never calls free(),
but it leaks memory like mad :)

> but even if you kill -SEGV `pgrep this` (Segmentation fault (core
> dumped) the memory is getting freed anyway (presumably by the glorious
> kernel). which you can see dynamicly by typing top in the console.

Yes.  When a process terminates, the kernel dismantles and releases all
its 'mapped memory areas', including the heap where malloc()'ed memory
comes from.